A double sequence in ℝᵖ is a function X with domain ℕ × ℕ and values in ℝᵖ; its value at (m, n) is written xₘₙ and the whole function is written X = (xₘₙ). In the customary array the first index names the row and the second the column. An element x is a limit (or double limit) of X in case for each ε > 0 there is N(ε) ∈ ℕ such that ‖xₘₙ − x‖ < ε for all m, n ≥ N(ε). One then writes x = lim (xₘₙ).
雙重數列就是把下標換成一對自然數，可以排成一張無窮的表格。雙重極限要求的是：表格右下角那一整塊（兩個下標都夠大）全部落進以 x 為心的 ε 球裡。
正例：xₘₙ = 1/(m + n) 的雙重極限是 0——m, n ≥ N 時 1/(m + n) ≤ 1/(2N)，取 N 夠大即可。反例：xₘₙ = m/(m + n)沒有雙重極限——不論 N 取多大，右下角那塊裡既有 x_{N,N²}（接近 0）也有 x_{N², N}（接近 1）。兩個下標同時夠大，並不表示它們的相對大小受到控制。
這張圖在說 19.4 要求的是什麼：不是某一列、也不是某一行，而是右下角那一整塊都得落進 ε 球。這個要求比「對角線收斂」強（把 m 與 n 都取成同一個 k ≥ N 即得）；與「每一列各自收斂」則互不蘊涵——後面兩篇的例子會把這些差別一一拆開。
例 6xₘₙ = 1/(m + n)
最簡單的一張表長什麼樣？
當 m, n ≥ N 時 m + n ≥ 2N，所以 0 < 1/(m + n) ≤ 1/(2N)。給定 ε > 0，由 §6-3 的 6.7 取 N 使 1/(2N) < ε，右下角那一整塊就全部落進 ε 之內。所以雙重極限是 0。
A double sequence X = (xₘₙ) in ℝᵖ is convergent if and only if for each ε > 0 there is M(ε) ∈ ℕ such that ‖xₘₙ − x_{rs}‖ < ε for all m, n, r, s ≥ M(ε).
正例：xₘₙ = 1/(m + n) 滿足判準——m, n, r, s ≥ M 時兩個值都落在 (0, 1/(2M)] 裡，相差不超過 1/(2M)。反例：條件裡的四個下標不能減成兩個——若只要求「‖x_{kk} − x_{ll}‖ 對夠大的 k, l 很小」（也就是只看對角線），xₘₙ = m/(m + n) 就會過關（對角線恆為 1/2），可是它沒有雙重極限。
PROOF
本讀本補上這條的證明。正方向是一次三角不等式；反方向的難處在於手上沒有候選極限，得先造一個出來。
證明計畫 · 由所求想起 正方向：所求是兩個元素靠得近。各自繞經極限 x，各分 ε/2。 反方向：所求是一個極限。沿對角線取出普通的數列(x_{kk})——四個下標同時取成 k 與 l 時，判準正好說它是 Cauchy 數列，完備性由 16.10 提供，極限就這樣造出來。最後把「離對角線上的點很近」過極限成「離 x 很近」。
Proof. Suppose x = lim (xₘₙ) and let ε > 0. Take M(ε) = N(ε/2) from Definition 19.4. If m, n, r, s ≥ M(ε), then ‖xₘₙ − x_{rs}‖ ≤ ‖xₘₙ − x‖ + ‖x − x_{rs}‖ < ε/2 + ε/2 = ε.
Conversely, assume the stated condition. Given ε > 0 and k, l ≥ M(ε), apply it with (m, n) = (k, k) and (r, s) = (l, l) to get ‖x_{kk} − x_{ll}‖ < ε. Thus the diagonal sequence (x_{kk} : k ∈ ℕ) is a Cauchy sequence in ℝᵖ, and by 16.10 it converges to some x ∈ ℝᵖ.
Let ε > 0 and put M = M(ε/2). Fix m, n ≥ M. For every k ≥ M the hypothesis gives ‖xₘₙ − x_{kk}‖ < ε/2. The real sequence aₖ = ‖xₘₙ − x_{kk}‖ converges to ‖xₘₙ − x‖, since |aₖ − ‖xₘₙ − x‖| ≤ ‖x_{kk} − x‖ → 0. Hence 15.8 gives ‖xₘₙ − x‖ ≤ ε/2 < ε, and therefore x = lim (xₘₙ).
Regard the m-th row of the array as the sequence Yₘ = (xₘₙ : n ∈ ℕ). Suppose each Yₘ converges, say to yₘ, and that the sequence (yₘ) converges. Its limit is called the row iterated limit of X and is written limₘ limₙ (xₘₙ). Regarding the n-th column as Zₙ = (xₘₙ : m ∈ ℕ) with limits zₙ and proceeding in the same way gives the column iterated limitlimₙ limₘ (xₘₙ).